[tex]P=\frac{16\sqrt{xy}}{x+y}+\frac{x^{2}+y^{2}}{xy}=\frac{16\sqrt{xy}}{x+y}+\frac{(x+y)^2}{xy}-2=\frac{8\sqrt{xy}}{x+y}+\frac{8\sqrt{xy}}{x+y}+\frac{(x+y)^2}{xy}-2\geq 3\sqrt[3]{\frac{8\sqrt{xy}}{x+y}.\frac{8\sqrt{xy}}{x+y}.\frac{(x+y)^2}{xy}}-2=3.4-2=10[/tex]
Dấu "=" xảy ra khi x = y.